3.1.39 \(\int \frac {1}{\sqrt {b \tan ^4(c+d x)}} \, dx\) [39]

Optimal. Leaf size=51 \[ -\frac {\tan (c+d x)}{d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{\sqrt {b \tan ^4(c+d x)}} \]

[Out]

-tan(d*x+c)/d/(tan(d*x+c)^4*b)^(1/2)-x*tan(d*x+c)^2/(tan(d*x+c)^4*b)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \begin {gather*} -\frac {\tan (c+d x)}{d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{\sqrt {b \tan ^4(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Tan[c + d*x]^4],x]

[Out]

-(Tan[c + d*x]/(d*Sqrt[b*Tan[c + d*x]^4])) - (x*Tan[c + d*x]^2)/Sqrt[b*Tan[c + d*x]^4]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {b \tan ^4(c+d x)}} \, dx &=\frac {\tan ^2(c+d x) \int \cot ^2(c+d x) \, dx}{\sqrt {b \tan ^4(c+d x)}}\\ &=-\frac {\tan (c+d x)}{d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan ^2(c+d x) \int 1 \, dx}{\sqrt {b \tan ^4(c+d x)}}\\ &=-\frac {\tan (c+d x)}{d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{\sqrt {b \tan ^4(c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.06, size = 43, normalized size = 0.84 \begin {gather*} -\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\tan ^2(c+d x)\right ) \tan (c+d x)}{d \sqrt {b \tan ^4(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Tan[c + d*x]^4],x]

[Out]

-((Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2]*Tan[c + d*x])/(d*Sqrt[b*Tan[c + d*x]^4]))

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Maple [A]
time = 0.07, size = 40, normalized size = 0.78

method	result	size
derivativedivides	\(-\frac {\tan \left (d x +c \right ) \left (\arctan \left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )+1\right )}{d \sqrt {b \left (\tan ^{4}\left (d x +c \right )\right )}}\)	\(40\)
default	\(-\frac {\tan \left (d x +c \right ) \left (\arctan \left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )+1\right )}{d \sqrt {b \left (\tan ^{4}\left (d x +c \right )\right )}}\)	\(40\)
risch	\(\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} x}{\sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {2 i \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{\sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} d}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(d*x+c)^4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/d*tan(d*x+c)*(arctan(tan(d*x+c))*tan(d*x+c)+1)/(b*tan(d*x+c)^4)^(1/2)

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Maxima [A]
time = 0.49, size = 27, normalized size = 0.53 \begin {gather*} -\frac {\frac {d x + c}{\sqrt {b}} + \frac {1}{\sqrt {b} \tan \left (d x + c\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(tan(d*x+c)^4*b)^(1/2),x, algorithm="maxima")

[Out]

-((d*x + c)/sqrt(b) + 1/(sqrt(b)*tan(d*x + c)))/d

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Fricas [A]
time = 0.37, size = 39, normalized size = 0.76 \begin {gather*} -\frac {\sqrt {b \tan \left (d x + c\right )^{4}} {\left (d x \tan \left (d x + c\right ) + 1\right )}}{b d \tan \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(tan(d*x+c)^4*b)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(b*tan(d*x + c)^4)*(d*x*tan(d*x + c) + 1)/(b*d*tan(d*x + c)^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b \tan ^{4}{\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(tan(d*x+c)**4*b)**(1/2),x)

[Out]

Integral(1/sqrt(b*tan(c + d*x)**4), x)

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Giac [A]
time = 0.54, size = 45, normalized size = 0.88 \begin {gather*} -\frac {\frac {2 \, {\left (d x + c\right )}}{\sqrt {b}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {b}} + \frac {1}{\sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(tan(d*x+c)^4*b)^(1/2),x, algorithm="giac")

[Out]

-1/2*(2*(d*x + c)/sqrt(b) - tan(1/2*d*x + 1/2*c)/sqrt(b) + 1/(sqrt(b)*tan(1/2*d*x + 1/2*c)))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {b\,{\mathrm {tan}\left (c+d\,x\right )}^4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(c + d*x)^4)^(1/2),x)

[Out]

int(1/(b*tan(c + d*x)^4)^(1/2), x)

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